I require assistance in doing the proofs for the two following questions.
Firstly, $\sin(3a) = 3\sin(a) - 4\sin^3(a)$
And secondly, I must reprove this using the formulae $\cos(n\theta) = (z^n + z^{-n})/2 $ and $\sin(n\theta) = (z^n - z^{-n})/2i$.
So far, I have used De Moivre's Theorem and then utilised Binomial Expansion and then equated my reals and imaginaries. But I do not know whether I have satisfied that task correctly.
By de Moivre's theorem, we have
$$(\cos x + i \sin x)^3=(\cos 3x + i \sin 3x)$$
Expand the term on the left hand side using the binomial theorem, $$(\cos 3x + i \sin 3x) =(\cos x + i \sin x)^3= \cos^3 x +3i\cos^2 x \sin x - 3\sin^2 x \cos x - i \sin^3 x$$
Now equating the real and imaginary part we have
$$\cos 3x=\cos^3x-3\sin^2x\cos x \qquad \text{and} \qquad \sin 3x=3\cos^2 x \sin x-\sin^3 x$$
Therefore $$\sin 3x=3\cos^2 x \sin x-\sin^3 x= 3(1-\sin^2 x)\sin x-\sin^3 x=3\sin x-4\sin^3 x$$ (using $\cos^2 x + \sin^2 x=1$)